Egg Drop Ideas: Solve the Physics Puzzle

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Homework Help Overview

The discussion revolves around a physics project involving an egg drop challenge, where the goal is to design a device that can catch an egg dropped from a height of 9.91 meters without breaking it. The participants are tasked with exploring concepts related to forces, impulses, momentum, and energy calculations associated with the drop.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster considers using materials like tissue paper or packing material to cushion the egg and increase the drop time to reduce impact force. They also attempt to calculate the final velocity of the egg using kinematic equations but express uncertainty about their results.
  • Some participants question the accuracy of the original poster's calculations and suggest reviewing the relevant equations. Others propose alternative ideas, such as using a bucket of water to catch the egg.
  • There is a discussion about the strength of the egg and the importance of deceleration time in the design of the catching device.

Discussion Status

The conversation includes various attempts to clarify the physics involved in the egg drop challenge. Some participants provide constructive feedback on the calculations and suggest further exploration of the equations. There is an ongoing exchange of ideas regarding potential designs and materials for the device, with no explicit consensus reached yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which requires them to discuss and analyze the physics concepts involved without providing complete solutions. There is an emphasis on understanding the underlying principles rather than simply arriving at a final answer.

Maddyk
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Homework Statement


I need help with egg drop ideas! For this project, I have to create something that will catch an egg (about 62.5 g) dropped from a height of 9.91 meters so that it will not break or crack. I also have to discuss forces, impulses, momentum, time and velocity changes of the egg. I also have to calculate the work done by my device.

Homework Equations


p=mv
Ft=p2-p1
V=Vo+at
d=do+Vo+.5at
V^2=V(o)^2+2a(d-do)
W=Fd

The Attempt at a Solution


Well, for my contraption I know I need to increase the time the egg drops in order to decrease the force impacted on its shell. I was thinking of putting tissue paper in a box or even getting the tiny packing material and putting it in a box? I just wanted to know if anyone had any other solutions
As for solving for final velocity, I used V^2=V(o)^2+2a(d-do)=0+2(-9.8 m/s^2)(-9.91 meters)=194 m/s
I don't think that's right though…
As for the other equations, i can't solve for them until I decide what I will be making my apparatus from...
 
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Maddyk said:
I don't think that's right though…
It's not. You might want to check your "relevant equations" for starters.
 
Bystander said:
It's not. You might want to check your "relevant equations" for starters.
Okay. I went over my notes again, and i used d=vt+1/2at^2 to solve for t (t=(2d/g)^1/2) being that g=a=-9.8, and i got 1.42 s for the time. Would I be able to plug that time into the Vf=Vo+at equation (a=-9.8) to get final velocity? I did that and got -13.9 m/s
 
Much better.
 
To catch an egg? A bucket of water.
 
Hello Maddyk, and welcome to PF :smile: !

An egg is incredibly strong. Just try to squeeze one by applying manual pressure from all around (if you hand is big enough). No chance. But a prick with a pin will easily go through (*).

Here you can see eggs are strong enough to walk on.

Well, for my contraption I know I need to increase the time the egg drops in order to decrease the force impacted on its shell.
You mean you want to increase the time the egg decelerates from this 14 m/s down to 0 m/s. Douq's idea seems brilliant to me (*), but from your problem statement I conclude you have to do some more work. Like finding out how much deceleration an egg can withstand.

As for the amount of work: Actually your " V2=V(o)2+2a(d-do)=0+2(-9.8 m/s2)(-9.91 meters) " is correct. So at landing time you have V2=194 m2/s2 , which is not the same as V = 194 m/s.

Bystander's advice still deserves following up: go through your equations (dimensions too) and check if they are correct.​

If you understand where your V2 equation comes from, you'll know straightaway how much work your contraption has to do !
And since work can be used to lift a weight, perhaps you can even design your device to recover a nice fraction of the kinetic energy by letting the egg fall on an oversized cloth that is draped over the bucket with suitable weights hanging from the corners. See how far the weights can be lifted and check with mgh of the egg!

Have fun :smile: !

(*) [edit] I forgot: so the area that catches the egg is important, and with a water surface you are exploiting that.
 

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